Stirling's Formula

Theorem

The factorial function can be approximated by the formula:

$n! \sim \sqrt {2 \pi n} \paren {\dfrac n e}^n$

where $\sim$ denotes asymptotically equal.

Refinement

A refinement of Stirling's Formula is:

$n! \sim \sqrt {2 \pi n} \paren {\dfrac n e}^n \paren {1 + \dfrac 1 {12 n} }$

Proof 1

Let $a_n = \dfrac {n!} {\sqrt {2 n} \paren {\frac n e}^n}$.

Part 1

It will be shown that:

$\lim_{n \mathop \to \infty} a_n = a$

for some constant $a$.

This will imply that:

$\ds \lim_{n \mathop \to \infty} \frac {n!} {a \sqrt{2 n} \paren {\frac n e}^n} = 1$

By applying Power Series Expansion for $\map \ln {1 + x}$:

 $\ds \map \ln {\frac {n + 1} n}$ $=$ $\ds \map \ln {n + 1 } - \map \ln n$ Difference of Logarithms $\ds$ $=$ $\ds \map \ln {n + 1 } + \map \ln 2 - \map \ln 2 - \map \ln n$ adding $0$ $\ds$ $=$ $\ds \map \ln {2n + 2 } - \map \ln {2n}$ Sum of Logarithms $\ds$ $=$ $\ds \map \ln {2n + 2 } - \map \ln {2n + 1 } + \map \ln {2n + 1 } - \map \ln {2n}$ adding $0$ $\ds$ $=$ $\ds \map \ln {\dfrac {2n + 2 }{2n + 1 } } - \map \ln {\dfrac {2n }{2n + 1 } }$ Difference of Logarithms $\ds$ $=$ $\ds \map \ln {1 + \frac 1 {2 n + 1} } - \map \ln {1 - \frac 1 {2 n + 1} }$ set $x = \dfrac 1 {2n + 1}$ $\ds$ $=$ $\ds \sum_{k \mathop = 1}^\infty \paren {-1}^{k - 1} \frac 1 k {\paren {\frac 1 {2 n + 1} }^k} - \paren {-\sum_{k \mathop = 1}^\infty \frac 1 k {\paren {\frac 1 {2 n + 1} }^k} }$ Power Series Expansion for Logarithm of 1 + x and Power Series Expansion for Logarithm of 1 + x/Corollary $\ds$ $=$ $\ds 2 \sum_{k \mathop = 0}^\infty \frac 1 {2 k + 1} \paren {\frac 1 {2 n + 1} }^{2 k + 1}$ even terms cancel out; odd terms doubled

Let $b_n = \ln a_n$.

 $\ds b_n$ $=$ $\ds \map \ln {\dfrac {n!} {\sqrt {2 n} \paren {\frac n e}^n} }$ $\ds$ $=$ $\ds \map \ln {n!} - \dfrac 1 2 \map \ln {2n} - n \map \ln {\frac n e }$ Sum of Logarithms and Difference of Logarithms $\ds b_n - b_{n + 1}$ $=$ $\ds \paren {\map \ln {n!} - \dfrac 1 2 \map \ln {2n} - n \map \ln {\frac n e } } - \paren {\map \ln {\paren {n + 1}!} - \dfrac 1 2 \map \ln {2n + 2} - \paren {n + 1} \map \ln {\frac {n + 1} e } }$ $\ds$ $=$ $\ds \map \ln {\dfrac {n!} {\paren {n + 1}!} } + \dfrac 1 2 \map \ln {\dfrac {2n + 2} {2n } } + n \map \ln {\frac {n + 1} n } + \map \ln {n + 1 } - \map \ln e$ Sum of Logarithms and Difference of Logarithms $\ds$ $=$ $\ds n \map \ln {\frac {n + 1} n } + \dfrac 1 2 \map \ln {\dfrac {n + 1} {n } } - \map \ln {n + 1 } + \map \ln {n + 1 } - \map \ln e$ rearranging $\ds$ $=$ $\ds \paren {n + \frac 1 2} \map \ln {\frac {n + 1} n} - 1$ $\ds$ $=$ $\ds \paren {n + \frac 1 2} 2 \sum_{k \mathop = 0}^\infty \frac 1 {2 k + 1} \paren {\frac 1 {2 n + 1} }^{2 k + 1} - 1$ substituting from above and Power Series Expansion for Logarithm of 1 + x $\ds$ $=$ $\ds \paren {2n + 1} \sum_{k \mathop = 0}^\infty \frac 1 {2 k + 1} \paren {\frac 1 {2 n + 1} }^{2 k + 1} - 1$ $\text {(1)}: \quad$ $\ds$ $=$ $\ds \sum_{k \mathop = 1}^\infty \frac 1 {2 k + 1} \paren {\frac 1 {2 n + 1} }^{2 k}$ canceling $k = 0$ from sum $\ds$ $>$ $\ds 0$ $\ds b_n$ $>$ $\ds b_{n + 1}$

so the sequence $\sequence {b_n}$ is (strictly) decreasing.

From $(1)$:

 $\ds b_n - b_{n + 1}$ $=$ $\ds \sum_{k \mathop = 1}^\infty \frac 1 {2 k + 1} \paren {\frac 1 {2 n + 1} }^{2 k}$ $\ds \leadsto \ \$ $\ds b_n - b_{n + 1}$ $<$ $\ds \sum_{k \mathop = 1}^\infty \paren {\frac 1 {2 n + 1} }^{2 k}$ as $\dfrac 1 {2 k + 1} < 1$ $\ds$ $=$ $\ds \dfrac {\paren {\dfrac 1 {2n + 1 } }^2 } {1 - \paren {\dfrac 1 {2n + 1 } }^2}$ Sum of Infinite Geometric Sequence $\ds$ $=$ $\ds \dfrac {\paren {\dfrac 1 {2n + 1 } }^2 } {1 - \paren {\dfrac 1 {2n + 1 } }^2} \dfrac {\paren {2n + 1 }^2} {\paren {2n + 1 }^2}$ multiplying by $1$ $\ds$ $=$ $\ds \dfrac 1 {\paren {2n + 1 }^2 - 1}$ $\ds$ $=$ $\ds \frac 1 {4 n \paren {n + 1} }$

Hence:

 $\ds b_1 - b_n$ $=$ $\ds \sum_{m \mathop = 1}^{n - 1} b_m - b_{m + 1}$ Definition of Telescoping Series $\ds$ $=$ $\ds \frac 1 4 \sum_{m \mathop = 1}^{n - 1} \frac 1 {m \paren {m + 1} }$ from above $\ds$ $<$ $\ds \frac 1 4 \sum_{m \mathop = 1}^\infty \frac 1 {m \paren {m + 1} }$ all the way to $\infty$ $\ds$ $=$ $\ds \frac 1 4$ Sum of Sequence of Products of Consecutive Reciprocals

and so:

 $\ds b_1$ $=$ $\ds \map \ln {1!} - \dfrac 1 2 \map \ln 2 - \map \ln {\frac 1 e }$ $\ds$ $=$ $\ds 1 - \dfrac 1 2 \map \ln 2$

and:

$b_1 - b_n < \dfrac 1 4$

Therefore:

$b_n > b_1 - \dfrac 1 4 = \dfrac 3 4 - \dfrac 1 2 \map \ln 2$

Thus by definition $\sequence {b_n}$ is bounded below.

By Monotone Convergence Theorem, it follows that $\sequence {b_n}$ is convergent.

Let $b$ denote its limit.

Then:

$\ds \lim_{n \mathop \to \infty} a_n = e^{\lim_{n \mathop \to \infty} b_n} = e^b = a$

as required.

$\Box$

Part 2

By Wallis's Product, we have:

 $\ds \frac \pi 2$ $=$ $\ds \prod_{k \mathop = 1}^\infty \frac {2 k} {2 k - 1} \cdot \frac {2 k} {2 k + 1}$ $\ds$ $=$ $\ds \prod \paren {\frac 2 1 \cdot \frac 2 3 } \paren {\frac 4 3 \cdot \frac 4 5 } \paren {\frac 6 5 \cdot \frac 6 7 } \cdots$ $\ds$ $=$ $\ds \lim_{n \mathop \to \infty} \dfrac 1 {2n + 1 } \prod_{k \mathop = 1}^n \frac {\paren {2k}^2} {\paren {2 k - 1 }^2 }$ $\ds$ $=$ $\ds \lim_{n \mathop \to \infty} \dfrac 1 {2n + 1 } \prod_{k \mathop = 1}^n \frac {\paren {2k}^2} {\paren {2 k - 1 }^2 } \dfrac {\paren {2k}^2} {\paren {2k}^2}$ multiplying by $1$ $\ds$ $=$ $\ds \lim_{n \mathop \to \infty} \dfrac 1 {2n + 1 } \prod_{k \mathop = 1}^n \frac {2^4 k^4} {\paren {\paren {2 k - 1 } \paren {2 k } }^2 }$ $\text {(2)}: \quad$ $\ds$ $=$ $\ds \lim_{n \mathop \to \infty} \frac {2^{4 n} \paren {n!}^4} {\paren {\paren {2 n}!}^2 \paren {2 n + 1} }$

In Part 1 it was proved that:

$n! \sim a \sqrt {2 n} \paren {\dfrac n e}^n$

Substituting for $n!$ in $(2)$ yields:

 $\ds \frac \pi 2$ $=$ $\ds \lim_{n \mathop \to \infty} \frac {2^{4 n} \paren {n!}^4} {\paren {\paren {2 n}!}^2 \paren {2 n + 1} }$ $\ds$ $=$ $\ds \lim_{n \mathop \to \infty} \frac {2^{4 n} \paren {a \sqrt {2 n} \paren {\dfrac n e}^n }^4 } {\paren {\paren {a \sqrt {4 n} \paren {\dfrac {2n} e}^{2n} } }^2 \paren {2 n + 1} }$ $\ds$ $=$ $\ds \lim_{n \mathop \to \infty} \frac {2^{4 n} a^4 \paren {4 n^2} n^{4 n} e^{-4 n} } {a^2 \paren {4 n} 2^{4 n} n^{4 n} e^{-4 n} \paren {2 n + 1} }$ Power of Power $\ds$ $=$ $\ds a^2 \lim_{n \mathop \to \infty} \frac n {2 n + 1}$ as most of it cancels out $\ds$ $=$ $\ds \frac {a^2} 2$ $\ds \leadsto \ \$ $\ds a$ $=$ $\ds \sqrt \pi$

Hence the result.

$\blacksquare$

Proof 2

Consider the sequence $\sequence {d_n}$ defined as:

$d_n = \map \ln {n!} - \paren {n + \dfrac 1 2} \ln n + n$

From Lemma 2 it is seen that $\sequence {d_n}$ is a decreasing sequence.

From Lemma 3 it is seen that the sequence:

$\sequence {d_n - \dfrac 1 {12 n} }$

is increasing.

In particular:

$\forall n \in \N_{>0}: d_n - \dfrac 1 {12 n} \ge d_1 = \dfrac 1 {12}$

and so $\sequence {d_n}$ is bounded below.

From the Monotone Convergence Theorem (Real Analysis), it follows that $\sequence {d_n}$ is convergent.

Let $d_n \to d$ as $n \to \infty$.

From Exponential Function is Continuous, we have:

$\exp d_n \to \exp d$ as $n \to \infty$

Let $C = \exp d$.

Then:

$\dfrac {n!} {n^n \sqrt n e^{-n} } \to C$ as $n \to \infty$

It remains to be shown that $C = \sqrt {2 \pi}$.

Let $I_n$ be defined as:

$\ds I_n = \int_0^{\frac \pi 2} \sin^n x \rd x$

Then from Lemma 4:

$\ds \lim_{n \mathop \to \infty} \frac {I_{2 n} } {I_{2 n + 1} } = 1$

So:

 $\ds \lim_{n \mathop \to \infty} \dfrac {n!} {n^n \sqrt n e^{-n} }$ $=$ $\ds \sqrt {2 \pi}$ Lemma 5 $\ds \leadsto \ \$ $\ds \lim_{n \mathop \to \infty} \dfrac {n!} {\sqrt {2 \pi n} \paren {\frac n e}^n}$ $=$ $\ds 1$

Hence the result.

$\blacksquare$

Also defined as

This result can also be seen reported as:

$n! \sim \sqrt {2 \pi} n^n n^{1/2} e^{-n}$

Other variants are sometimes encountered.

Also known as

Stirling's Formula is otherwise known as Stirling's approximation.

Examples

Factorial of $8$

The factorial of $8$ is given by Stirling's Formula as:

$8! \approx 39 \ 902$

which shows an error of about $1 \%$.

Source of Name

This entry was named for James Stirling.